Python 实用宝典

Question 1

How do you check whether a file is a normal file or a directory using python?

Question 2

os.path.isdir() and os.path.isfile() should give you what you want. See: http://docs.python.org/library/os.path.html

Question 3

As other answers have said, os.path.isdir() and os.path.isfile() are what you want. However, you need to keep in mind that these are not the only two cases. Use os.path.islink() for symlinks for instance. Furthermore, these all return False if the file does not exist, so you’ll probably want to check with os.path.exists() as well.

Question 4

Python 3.4 introduced the pathlib module into the standard library, which provides an object oriented approach to handle filesystem paths. The relavant methods would be .is_file() and .is_dir():

In [1]: from pathlib import Path

In [2]: p = Path('/usr')

In [3]: p.is_file()
Out[3]: False

In [4]: p.is_dir()
Out[4]: True

In [5]: q = p / 'bin' / 'vim'

In [6]: q.is_file()
Out[6]: True

In [7]: q.is_dir()
Out[7]: False

Pathlib is also available on Python 2.7 via the pathlib2 module on PyPi.

Question 5

import os

if os.path.isdir(d):
    print "dir"
else:
    print "file"

Question 6

os.path.isdir('string')
os.path.isfile('string')

Question 7

try this:

import os.path
if os.path.isdir("path/to/your/file"):
    print "it's a directory"
else:
    print "it's a file"

Question 8

If you’re just stepping through a set of directories you might be better just to try os.chdir and give an error/warning if it fails:

import os,sys
for DirName in sys.argv[1:]:
    SaveDir = os.getcwd()
    try:
        os.chdir(DirName)
        print "Changed to "+DirName
        # Do some stuff here in the directory
        os.chdir(SaveDir)
    except:
        sys.stderr.write("%s: WARNING: Cannot change to %s\n" % (sys.argv[0],DirName))

Question 9

From this page, we know that:

Chained comparisons are faster than using the and operator. Write x < y < z instead of x < y and y < z.

However, I got a different result testing the following code snippets:

$ python -m timeit "x = 1.2" "y = 1.3" "z = 1.8" "x < y < z"
1000000 loops, best of 3: 0.322 usec per loop
$ python -m timeit "x = 1.2" "y = 1.3" "z = 1.8" "x < y and y < z"
1000000 loops, best of 3: 0.22 usec per loop
$ python -m timeit "x = 1.2" "y = 1.3" "z = 1.1" "x < y < z"
1000000 loops, best of 3: 0.279 usec per loop
$ python -m timeit "x = 1.2" "y = 1.3" "z = 1.1" "x < y and y < z"
1000000 loops, best of 3: 0.215 usec per loop

It seems that x < y and y < z is faster than x < y < z. Why?

After searching some posts in this site (like this one) I know that “evaluated only once” is the key for x < y < z, however I’m still confused. To do further study, I disassembled these two functions using dis.dis:

import dis

def chained_compare():
        x = 1.2
        y = 1.3
        z = 1.1
        x < y < z

def and_compare():
        x = 1.2
        y = 1.3
        z = 1.1
        x < y and y < z

dis.dis(chained_compare)
dis.dis(and_compare)

And the output is:

## chained_compare ##

  4           0 LOAD_CONST               1 (1.2)
              3 STORE_FAST               0 (x)

  5           6 LOAD_CONST               2 (1.3)
              9 STORE_FAST               1 (y)

  6          12 LOAD_CONST               3 (1.1)
             15 STORE_FAST               2 (z)

  7          18 LOAD_FAST                0 (x)
             21 LOAD_FAST                1 (y)
             24 DUP_TOP
             25 ROT_THREE
             26 COMPARE_OP               0 (<)
             29 JUMP_IF_FALSE_OR_POP    41
             32 LOAD_FAST                2 (z)
             35 COMPARE_OP               0 (<)
             38 JUMP_FORWARD             2 (to 43)
        >>   41 ROT_TWO
             42 POP_TOP
        >>   43 POP_TOP
             44 LOAD_CONST               0 (None)
             47 RETURN_VALUE

## and_compare ##

 10           0 LOAD_CONST               1 (1.2)
              3 STORE_FAST               0 (x)

 11           6 LOAD_CONST               2 (1.3)
              9 STORE_FAST               1 (y)

 12          12 LOAD_CONST               3 (1.1)
             15 STORE_FAST               2 (z)

 13          18 LOAD_FAST                0 (x)
             21 LOAD_FAST                1 (y)
             24 COMPARE_OP               0 (<)
             27 JUMP_IF_FALSE_OR_POP    39
             30 LOAD_FAST                1 (y)
             33 LOAD_FAST                2 (z)
             36 COMPARE_OP               0 (<)
        >>   39 POP_TOP
             40 LOAD_CONST               0 (None)

It seems that the x < y and y < z has less dissembled commands than x < y < z. Should I consider x < y and y < z faster than x < y < z?

Tested with Python 2.7.6 on an Intel(R) Xeon(R) CPU E5640 @ 2.67GHz.

Question 10

The difference is that in x < y < z y is only evaluated once. This does not make a large difference if y is a variable, but it does when it is a function call, which takes some time to compute.

from time import sleep
def y():
    sleep(.2)
    return 1.3
%timeit 1.2 < y() < 1.8
10 loops, best of 3: 203 ms per loop
%timeit 1.2 < y() and y() < 1.8
1 loops, best of 3: 405 ms per loop

Question 11

Optimal bytecode for both of the functions you defined would be

          0 LOAD_CONST               0 (None)
          3 RETURN_VALUE

because the result of the comparison is not used. Let’s make the situation more interesting by returning the result of the comparison. Let’s also have the result not be knowable at compile time.

def interesting_compare(y):
    x = 1.1
    z = 1.3
    return x < y < z  # or: x < y and y < z

Again, the two versions of the comparison are semantically identical, so the optimal bytecode is the same for both constructs. As best I can work it out, it would look like this. I’ve annotated each line with the stack contents before and after each opcode, in Forth notation (top of stack at right, -- divides before and after, trailing ? indicates something that might or might not be there). Note that RETURN_VALUE discards everything that happens to be left on the stack underneath the value returned.

          0 LOAD_FAST                0 (y)    ;          -- y
          3 DUP_TOP                           ; y        -- y y
          4 LOAD_CONST               0 (1.1)  ; y y      -- y y 1.1
          7 COMPARE_OP               4 (>)    ; y y 1.1  -- y pred
         10 JUMP_IF_FALSE_OR_POP     19       ; y pred   -- y
         13 LOAD_CONST               1 (1.3)  ; y        -- y 1.3
         16 COMPARE_OP               0 (<)    ; y 1.3    -- pred
     >>  19 RETURN_VALUE                      ; y? pred  --

If an implementation of the language, CPython, PyPy, whatever, does not generate this bytecode (or its own equivalent sequence of operations) for both variations, that demonstrates the poor quality of that bytecode compiler. Getting from the bytecode sequences you posted to the above is a solved problem (I think all you need for this case is constant folding, dead code elimination, and better modeling of the contents of the stack; common subexpression elimination would also be cheap and valuable), and there’s really no excuse for not doing it in a modern language implementation.

Now, it happens that all current implementations of the language have poor-quality bytecode compilers. But you should ignore that while coding! Pretend the bytecode compiler is good, and write the most readable code. It will probably be plenty fast enough anyway. If it isn’t, look for algorithmic improvements first, and give Cython a try second — that will provide far more improvement for the same effort than any expression-level tweaks you might apply.

Question 12

Since the difference in the output seem to be due to lack of optimization I think you should ignore that difference for most cases – it could be that the difference will go away. The difference is because y only should be evaluated once and that is solved by duplicating it on the stack which requires an extra POP_TOP – the solution to use LOAD_FAST might be possible though.

The important difference though is that in x<y and y<z the second y should be evaluated twice if x<y evaluates to true, this has implications if the evaluation of y takes considerable time or have side effects.

In most scenarios you should use x<y<z despite the fact it’s somewhat slower.

Question 13

First of all, your comparison is pretty much meaningless because the two different constructs were not introduced to provide a performance improvement, so you shouldn’t decide whether to use one in place of the other based on that.

The x < y < z construct:

Is clearer and more direct in its meaning.
Its semantics is what you’d expect from the “mathematical meaning” of the comparison: evalute x, y and z once and check if the whole condition holds. Using and changes the semantics by evaluating y multiple times, which can change the result.

So choose one in place of the other depending on the semantics you want and, if they are equivalent, whether one is more readable than the other.

This said: more disassembled code does does not imply slower code. However executing more bytecode operations means that each operation is simpler and yet it requires an iteration of the main loop. This means that if the operations you are performing are extremely fast (e.g. local variable lookup as you are doing there), then the overhead of executing more bytecode operations can matter.

But note that this result does not hold in the more generic situation, only to the “worst case” that you happen to profile. As others have noted, if you change y to something that takes even a bit more time you’ll see that the results change, because the chained notation evaluates it only once.

Summarizing:

Consider semantics before performance.
Take into account readability.
Don’t trust micro benchmarks. Always profile with different kind of parameters to see how a function/expression timing behave in relation to said parameters and consider how you plan to use it.

Question 14

If I define a class method with a keyword argument thus:

class foo(object):
  def foodo(thing=None, thong='not underwear'):
    print thing if thing else "nothing" 
    print 'a thong is',thong

calling the method generates a TypeError:

myfoo = foo()
myfoo.foodo(thing="something")

...
TypeError: foodo() got multiple values for keyword argument 'thing'

What’s going on?

Question 15

The problem is that the first argument passed to class methods in python is always a copy of the class instance on which the method is called, typically labelled self. If the class is declared thus:

class foo(object):
  def foodo(self, thing=None, thong='not underwear'):
    print thing if thing else "nothing" 
    print 'a thong is',thong

it behaves as expected.

Explanation:

Without self as the first parameter, when myfoo.foodo(thing="something") is executed, the foodo method is called with arguments (myfoo, thing="something"). The instance myfoo is then assigned to thing (since thing is the first declared parameter), but python also attempts to assign "something" to thing, hence the Exception.

To demonstrate, try running this with the original code:

myfoo.foodo("something")
print
print myfoo

You’ll output like:

<__main__.foo object at 0x321c290>
a thong is something

<__main__.foo object at 0x321c290>

You can see that ‘thing’ has been assigned a reference to the instance ‘myfoo’ of the class ‘foo’. This section of the docs explains how function arguments work a bit more.

Question 16

Thanks for the instructive posts. I’d just like to keep a note that if you’re getting “TypeError: foodo() got multiple values for keyword argument ‘thing'”, it may also be that you’re mistakenly passing the ‘self’ as a parameter when calling the function (probably because you copied the line from the class declaration – it’s a common error when one’s in a hurry).

Question 17

This might be obvious, but it might help someone who has never seen it before. This also happens for regular functions if you mistakenly assign a parameter by position and explicitly by name.

>>> def foodo(thing=None, thong='not underwear'):
...     print thing if thing else "nothing"
...     print 'a thong is',thong
...
>>> foodo('something', thing='everything')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foodo() got multiple values for keyword argument 'thing'

Question 18

just add ‘staticmethod’ decorator to function and problem is fixed

class foo(object):
    @staticmethod
    def foodo(thing=None, thong='not underwear'):
        print thing if thing else "nothing" 
        print 'a thong is',thong

Question 19

I want to add one more answer :

It happens when you try to pass positional parameter with wrong position order along with keyword argument in calling function.

there is difference between parameter and argument you can read in detail about here Arguments and Parameter in python

def hello(a,b=1, *args):
   print(a, b, *args)


hello(1, 2, 3, 4,a=12)

since we have three parameters :

a is positional parameter

b=1 is keyword and default parameter

*args is variable length parameter

so we first assign a as positional parameter , means we have to provide value to positional argument in its position order, here order matter. but we are passing argument 1 at the place of a in calling function and then we are also providing value to a , treating as keyword argument. now a have two values :

one is positional value: a=1

second is keyworded value which is a=12

Solution

We have to change hello(1, 2, 3, 4,a=12) to hello(1, 2, 3, 4,12) so now a will get only one positional value which is 1 and b will get value 2 and rest of values will get *args (variable length parameter)

additional information

if we want that *args should get 2,3,4 and a should get 1 and b should get 12

then we can do like this
def hello(a,*args,b=1): pass hello(1, 2, 3, 4,b=12)

Something more :

def hello(a,*c,b=1,**kwargs):
    print(b)
    print(c)
    print(a)
    print(kwargs)

hello(1,2,1,2,8,9,c=12)

output :

1

(2, 1, 2, 8, 9)

1

{'c': 12}

Question 20

This error can also happen if you pass a key word argument for which one of the keys is similar (has same string name) to a positional argument.

>>> class Foo():
...     def bar(self, bar, **kwargs):
...             print(bar)
... 
>>> kwgs = {"bar":"Barred", "jokes":"Another key word argument"}
>>> myfoo = Foo()
>>> myfoo.bar("fire", **kwgs)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: bar() got multiple values for argument 'bar'
>>>

“fire” has been accepted into the ‘bar’ argument. And yet there is another ‘bar’ argument present in kwargs.

You would have to remove the keyword argument from the kwargs before passing it to the method.

Question 21

Also this can happen in Django if you are using jquery ajax to url that reverses to a function that doesn’t contain ‘request’ parameter

$.ajax({
  url: '{{ url_to_myfunc }}',
});


def myfunc(foo, bar):
    ...

Question 22

When I use subcommands with python argparse, I can get the selected arguments.

parser = argparse.ArgumentParser()
parser.add_argument('-g', '--global')
subparsers = parser.add_subparsers()   
foo_parser = subparsers.add_parser('foo')
foo_parser.add_argument('-c', '--count')
bar_parser = subparsers.add_parser('bar')
args = parser.parse_args(['-g, 'xyz', 'foo', '--count', '42'])
# args => Namespace(global='xyz', count='42')

So args doesn’t contain 'foo'. Simply writing sys.argv[1] doesn’t work because of the possible global args. How can I get the subcommand itself?

Question 23

The very bottom of the Python docs on argparse sub-commands explains how to do this:

>>> parser = argparse.ArgumentParser()
>>> parser.add_argument('-g', '--global')
>>> subparsers = parser.add_subparsers(dest="subparser_name") # this line changed
>>> foo_parser = subparsers.add_parser('foo')
>>> foo_parser.add_argument('-c', '--count')
>>> bar_parser = subparsers.add_parser('bar')
>>> args = parser.parse_args(['-g', 'xyz', 'foo', '--count', '42'])
>>> args
Namespace(count='42', global='xyz', subparser_name='foo')

You can also use the set_defaults() method referenced just above the example I found.

Question 24

ArgumentParser.add_subparsers has dest formal argument described as:

dest – name of the attribute under which sub-command name will be stored; by default None and no value is stored

In the example below of a simple task function layout using subparsers, the selected subparser is in parser.parse_args().subparser.

import argparse


def task_a(alpha):
    print('task a', alpha)


def task_b(beta, gamma):
    print('task b', beta, gamma)


if __name__ == '__main__':
    parser = argparse.ArgumentParser()
    subparsers = parser.add_subparsers(dest='subparser')

    parser_a = subparsers.add_parser('task_a')
    parser_a.add_argument(
        '-a', '--alpha', dest='alpha', help='Alpha description')

    parser_b = subparsers.add_parser('task_b')
    parser_b.add_argument(
        '-b', '--beta', dest='beta', help='Beta description')
    parser_b.add_argument(
        '-g', '--gamma', dest='gamma', default=42, help='Gamma description')

    kwargs = vars(parser.parse_args())
    globals()[kwargs.pop('subparser')](**kwargs)

Question 25

I was reading ‘Dive Into Python’ and in the chapter on classes it gives this example:

class FileInfo(UserDict):
    "store file metadata"
    def __init__(self, filename=None):
        UserDict.__init__(self)
        self["name"] = filename

The author then says that if you want to override the __init__ method, you must explicitly call the parent __init__ with the correct parameters.

What if that FileInfo class had more than one ancestor class?
- Do I have to explicitly call all of the ancestor classes’ __init__ methods?
Also, do I have to do this to any other method I want to override?

Question 26

The book is a bit dated with respect to subclass-superclass calling. It’s also a little dated with respect to subclassing built-in classes.

It looks like this nowadays:

class FileInfo(dict):
    """store file metadata"""
    def __init__(self, filename=None):
        super(FileInfo, self).__init__()
        self["name"] = filename

Note the following:

We can directly subclass built-in classes, like dict, list, tuple, etc.
The super function handles tracking down this class’s superclasses and calling functions in them appropriately.

Question 27

In each class that you need to inherit from, you can run a loop of each class that needs init’d upon initiation of the child class…an example that can copied might be better understood…

class Female_Grandparent:
    def __init__(self):
        self.grandma_name = 'Grandma'

class Male_Grandparent:
    def __init__(self):
        self.grandpa_name = 'Grandpa'

class Parent(Female_Grandparent, Male_Grandparent):
    def __init__(self):
        Female_Grandparent.__init__(self)
        Male_Grandparent.__init__(self)

        self.parent_name = 'Parent Class'

class Child(Parent):
    def __init__(self):
        Parent.__init__(self)
#---------------------------------------------------------------------------------------#
        for cls in Parent.__bases__: # This block grabs the classes of the child
             cls.__init__(self)      # class (which is named 'Parent' in this case), 
                                     # and iterates through them, initiating each one.
                                     # The result is that each parent, of each child,
                                     # is automatically handled upon initiation of the 
                                     # dependent class. WOOT WOOT! :D
#---------------------------------------------------------------------------------------#



g = Female_Grandparent()
print g.grandma_name

p = Parent()
print p.grandma_name

child = Child()

print child.grandma_name

Question 28

You don’t really have to call the __init__ methods of the base class(es), but you usually want to do it because the base classes will do some important initializations there that are needed for rest of the classes methods to work.

For other methods it depends on your intentions. If you just want to add something to the base classes behavior you will want to call the base classes method additionally to your own code. If you want to fundamentally change the behavior, you might not call the base class’ method and implement all the functionality directly in the derived class.

Question 29

If the FileInfo class has more than one ancestor class then you should definitely call all of their __init__() functions. You should also do the same for the __del__() function, which is a destructor.

Question 30

Yes, you must call __init__ for each parent class. The same goes for functions, if you are overriding a function that exists in both parents.

Question 31

How do you check whether a string contains only numbers?

I’ve given it a go here. I’d like to see the simplest way to accomplish this.

import string

def main():
    isbn = input("Enter your 10 digit ISBN number: ")
    if len(isbn) == 10 and string.digits == True:
        print ("Works")
    else:
        print("Error, 10 digit number was not inputted and/or letters were inputted.")
        main()

if __name__ == "__main__":
    main()
    input("Press enter to exit: ")

Question 32

You’ll want to use the isdigit method on your str object:

if len(isbn) == 10 and isbn.isdigit():

From the isdigit documentation:

str.isdigit()

Return True if all characters in the string are digits and there is at least one character, False otherwise. Digits include decimal characters and digits that need special handling, such as the compatibility superscript digits. This covers digits which cannot be used to form numbers in base 10, like the Kharosthi numbers. Formally, a digit is a character that has the property value Numeric_Type=Digit or Numeric_Type=Decimal.

Question 33

Use str.isdigit:

>>> "12345".isdigit()
True
>>> "12345a".isdigit()
False
>>>

Question 34

Use string isdigit function:

>>> s = '12345'
>>> s.isdigit()
True
>>> s = '1abc'
>>> s.isdigit()
False

Question 35

You can also use the regex,

import re

eg:-1) word = “3487954”

re.match('^[0-9]*$',word)

eg:-2) word = “3487.954”

re.match('^[0-9\.]*$',word)

eg:-3) word = “3487.954 328”

re.match('^[0-9\.\ ]*$',word)

As you can see all 3 eg means that there is only no in your string. So you can follow the respective solutions given with them.

Question 36

What about of float numbers, negatives numbers, etc.. All the examples before will be wrong.

Until now I got something like this, but I think it could be a lot better:

'95.95'.replace('.','',1).isdigit()

will return true only if there is one or no ‘.’ in the string of digits.

'9.5.9.5'.replace('.','',1).isdigit()

will return false

Question 37

As pointed out in this comment How do you check in python whether a string contains only numbers? the isdigit() method is not totally accurate for this use case, because it returns True for some digit-like characters:

>>> "\u2070".isdigit() # unicode escaped 'superscript zero' 
True

If this needs to be avoided, the following simple function checks, if all characters in a string are a digit between “0” and “9”:

import string

def contains_only_digits(s):
    # True for "", "0", "123"
    # False for "1.2", "1,2", "-1", "a", "a1"
    for ch in s:
        if not ch in string.digits:
            return False
    return True

Used in the example from the question:

if len(isbn) == 10 and contains_only_digits(isbn):
    print ("Works")

Question 38

You can use try catch block here:

s="1234"
try:
    num=int(s)
    print "S contains only digits"
except:
    print "S doesn't contain digits ONLY"

Question 39

As every time I encounter an issue with the check is because the str can be None sometimes, and if the str can be None, only use str.isdigit() is not enough as you will get an error

AttributeError: ‘NoneType’ object has no attribute ‘isdigit’

and then you need to first validate the str is None or not. To avoid a multi-if branch, a clear way to do this is:

if str and str.isdigit():

Hope this helps for people have the same issue like me.

Question 40

There are 2 methods that I can think of to check whether a string has all digits of not

Method 1(Using the built-in isdigit() function in python):-

>>>st = '12345'
>>>st.isdigit()
True
>>>st = '1abcd'
>>>st.isdigit()
False

Method 2(Performing Exception Handling on top of the string):-

st="1abcd"
try:
    number=int(st)
    print("String has all digits in it")
except:
    print("String does not have all digits in it")

The output of the above code will be:

String does not have all digits in it

Question 41

you can use str.isdigit() method or str.isnumeric() method

Question 42

I need to merge multiple dictionaries, here’s what I have for instance:

dict1 = {1:{"a":{A}}, 2:{"b":{B}}}

dict2 = {2:{"c":{C}}, 3:{"d":{D}}

With A B C and D being leaves of the tree, like {"info1":"value", "info2":"value2"}

There is an unknown level(depth) of dictionaries, it could be {2:{"c":{"z":{"y":{C}}}}}

In my case it represents a directory/files structure with nodes being docs and leaves being files.

I want to merge them to obtain:

 dict3 = {1:{"a":{A}}, 2:{"b":{B},"c":{C}}, 3:{"d":{D}}}

I’m not sure how I could do that easily with Python.

Question 43

this is actually quite tricky – particularly if you want a useful error message when things are inconsistent, while correctly accepting duplicate but consistent entries (something no other answer here does….)

assuming you don’t have huge numbers of entries a recursive function is easiest:

def merge(a, b, path=None):
    "merges b into a"
    if path is None: path = []
    for key in b:
        if key in a:
            if isinstance(a[key], dict) and isinstance(b[key], dict):
                merge(a[key], b[key], path + [str(key)])
            elif a[key] == b[key]:
                pass # same leaf value
            else:
                raise Exception('Conflict at %s' % '.'.join(path + [str(key)]))
        else:
            a[key] = b[key]
    return a

# works
print(merge({1:{"a":"A"},2:{"b":"B"}}, {2:{"c":"C"},3:{"d":"D"}}))
# has conflict
merge({1:{"a":"A"},2:{"b":"B"}}, {1:{"a":"A"},2:{"b":"C"}})

note that this mutates a – the contents of b are added to a (which is also returned). if you want to keep a you could call it like merge(dict(a), b).

agf pointed out (below) that you may have more than two dicts, in which case you can use:

reduce(merge, [dict1, dict2, dict3...])

where everything will be added to dict1.

[note – i edited my initial answer to mutate the first argument; that makes the “reduce” easier to explain]

ps in python 3, you will also need from functools import reduce

Question 44

Here’s an easy way to do it using generators:

def mergedicts(dict1, dict2):
    for k in set(dict1.keys()).union(dict2.keys()):
        if k in dict1 and k in dict2:
            if isinstance(dict1[k], dict) and isinstance(dict2[k], dict):
                yield (k, dict(mergedicts(dict1[k], dict2[k])))
            else:
                # If one of the values is not a dict, you can't continue merging it.
                # Value from second dict overrides one in first and we move on.
                yield (k, dict2[k])
                # Alternatively, replace this with exception raiser to alert you of value conflicts
        elif k in dict1:
            yield (k, dict1[k])
        else:
            yield (k, dict2[k])

dict1 = {1:{"a":"A"},2:{"b":"B"}}
dict2 = {2:{"c":"C"},3:{"d":"D"}}

print dict(mergedicts(dict1,dict2))

This prints:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}

Question 45

One issue with this question is that the values of the dict can be arbitrarily complex pieces of data. Based upon these and other answers I came up with this code:

class YamlReaderError(Exception):
    pass

def data_merge(a, b):
    """merges b into a and return merged result

    NOTE: tuples and arbitrary objects are not handled as it is totally ambiguous what should happen"""
    key = None
    # ## debug output
    # sys.stderr.write("DEBUG: %s to %s\n" %(b,a))
    try:
        if a is None or isinstance(a, str) or isinstance(a, unicode) or isinstance(a, int) or isinstance(a, long) or isinstance(a, float):
            # border case for first run or if a is a primitive
            a = b
        elif isinstance(a, list):
            # lists can be only appended
            if isinstance(b, list):
                # merge lists
                a.extend(b)
            else:
                # append to list
                a.append(b)
        elif isinstance(a, dict):
            # dicts must be merged
            if isinstance(b, dict):
                for key in b:
                    if key in a:
                        a[key] = data_merge(a[key], b[key])
                    else:
                        a[key] = b[key]
            else:
                raise YamlReaderError('Cannot merge non-dict "%s" into dict "%s"' % (b, a))
        else:
            raise YamlReaderError('NOT IMPLEMENTED "%s" into "%s"' % (b, a))
    except TypeError, e:
        raise YamlReaderError('TypeError "%s" in key "%s" when merging "%s" into "%s"' % (e, key, b, a))
    return a

My use case is merging YAML files where I only have to deal with a subset of possible data types. Hence I can ignore tuples and other objects. For me a sensible merge logic means

replace scalars
append lists
merge dicts by adding missing keys and updating existing keys

Everything else and the unforeseens results in an error.

Question 46

Dictionaries of dictionaries merge

As this is the canonical question (in spite of certain non-generalities) I’m providing the canonical Pythonic approach to solving this issue.

Simplest Case: “leaves are nested dicts that end in empty dicts”:

d1 = {'a': {1: {'foo': {}}, 2: {}}}
d2 = {'a': {1: {}, 2: {'bar': {}}}}
d3 = {'b': {3: {'baz': {}}}}
d4 = {'a': {1: {'quux': {}}}}

This is the simplest case for recursion, and I would recommend two naive approaches:

def rec_merge1(d1, d2):
    '''return new merged dict of dicts'''
    for k, v in d1.items(): # in Python 2, use .iteritems()!
        if k in d2:
            d2[k] = rec_merge1(v, d2[k])
    d3 = d1.copy()
    d3.update(d2)
    return d3

def rec_merge2(d1, d2):
    '''update first dict with second recursively'''
    for k, v in d1.items(): # in Python 2, use .iteritems()!
        if k in d2:
            d2[k] = rec_merge2(v, d2[k])
    d1.update(d2)
    return d1

I believe I would prefer the second to the first, but keep in mind that the original state of the first would have to be rebuilt from its origin. Here’s the usage:

>>> from functools import reduce # only required for Python 3.
>>> reduce(rec_merge1, (d1, d2, d3, d4))
{'a': {1: {'quux': {}, 'foo': {}}, 2: {'bar': {}}}, 'b': {3: {'baz': {}}}}
>>> reduce(rec_merge2, (d1, d2, d3, d4))
{'a': {1: {'quux': {}, 'foo': {}}, 2: {'bar': {}}}, 'b': {3: {'baz': {}}}}

Complex Case: “leaves are of any other type:”

So if they end in dicts, it’s a simple case of merging the end empty dicts. If not, it’s not so trivial. If strings, how do you merge them? Sets can be updated similarly, so we could give that treatment, but we lose the order in which they were merged. So does order matter?

So in lieu of more information, the simplest approach will be to give them the standard update treatment if both values are not dicts: i.e. the second dict’s value will overwrite the first, even if the second dict’s value is None and the first’s value is a dict with a lot of info.

d1 = {'a': {1: 'foo', 2: None}}
d2 = {'a': {1: None, 2: 'bar'}}
d3 = {'b': {3: 'baz'}}
d4 = {'a': {1: 'quux'}}

from collections.abc import MutableMapping

def rec_merge(d1, d2):
    '''
    Update two dicts of dicts recursively, 
    if either mapping has leaves that are non-dicts, 
    the second's leaf overwrites the first's.
    '''
    for k, v in d1.items():
        if k in d2:
            # this next check is the only difference!
            if all(isinstance(e, MutableMapping) for e in (v, d2[k])):
                d2[k] = rec_merge(v, d2[k])
            # we could further check types and merge as appropriate here.
    d3 = d1.copy()
    d3.update(d2)
    return d3

And now

from functools import reduce
reduce(rec_merge, (d1, d2, d3, d4))

returns

{'a': {1: 'quux', 2: 'bar'}, 'b': {3: 'baz'}}

Application to the original question:

I’ve had to remove the curly braces around the letters and put them in single quotes for this to be legit Python (else they would be set literals in Python 2.7+) as well as append a missing brace:

dict1 = {1:{"a":'A'}, 2:{"b":'B'}}
dict2 = {2:{"c":'C'}, 3:{"d":'D'}}

and rec_merge(dict1, dict2) now returns:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}

Which matches the desired outcome of the original question (after changing, e.g. the {A} to 'A'.)

Question 47

Based on @andrew cooke. This version handles nested lists of dicts and also allows the option to update the values

def merge(a, b, path=None, update=True):
    "http://stackoverflow.com/questions/7204805/python-dictionaries-of-dictionaries-merge"
    "merges b into a"
    if path is None: path = []
    for key in b:
        if key in a:
            if isinstance(a[key], dict) and isinstance(b[key], dict):
                merge(a[key], b[key], path + [str(key)])
            elif a[key] == b[key]:
                pass # same leaf value
            elif isinstance(a[key], list) and isinstance(b[key], list):
                for idx, val in enumerate(b[key]):
                    a[key][idx] = merge(a[key][idx], b[key][idx], path + [str(key), str(idx)], update=update)
            elif update:
                a[key] = b[key]
            else:
                raise Exception('Conflict at %s' % '.'.join(path + [str(key)]))
        else:
            a[key] = b[key]
    return a

Question 48

This simple recursive procedure will merge one dictionary into another while overriding conflicting keys:

#!/usr/bin/env python2.7

def merge_dicts(dict1, dict2):
    """ Recursively merges dict2 into dict1 """
    if not isinstance(dict1, dict) or not isinstance(dict2, dict):
        return dict2
    for k in dict2:
        if k in dict1:
            dict1[k] = merge_dicts(dict1[k], dict2[k])
        else:
            dict1[k] = dict2[k]
    return dict1

print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {2:{"c":"C"}, 3:{"d":"D"}}))
print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {1:{"a":"A"}, 2:{"b":"C"}}))

Output:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
{1: {'a': 'A'}, 2: {'b': 'C'}}

Question 49

Based on answers from @andrew cooke. It takes care of nested lists in a better way.

def deep_merge_lists(original, incoming):
    """
    Deep merge two lists. Modifies original.
    Recursively call deep merge on each correlated element of list. 
    If item type in both elements are
     a. dict: Call deep_merge_dicts on both values.
     b. list: Recursively call deep_merge_lists on both values.
     c. any other type: Value is overridden.
     d. conflicting types: Value is overridden.

    If length of incoming list is more that of original then extra values are appended.
    """
    common_length = min(len(original), len(incoming))
    for idx in range(common_length):
        if isinstance(original[idx], dict) and isinstance(incoming[idx], dict):
            deep_merge_dicts(original[idx], incoming[idx])

        elif isinstance(original[idx], list) and isinstance(incoming[idx], list):
            deep_merge_lists(original[idx], incoming[idx])

        else:
            original[idx] = incoming[idx]

    for idx in range(common_length, len(incoming)):
        original.append(incoming[idx])


def deep_merge_dicts(original, incoming):
    """
    Deep merge two dictionaries. Modifies original.
    For key conflicts if both values are:
     a. dict: Recursively call deep_merge_dicts on both values.
     b. list: Call deep_merge_lists on both values.
     c. any other type: Value is overridden.
     d. conflicting types: Value is overridden.

    """
    for key in incoming:
        if key in original:
            if isinstance(original[key], dict) and isinstance(incoming[key], dict):
                deep_merge_dicts(original[key], incoming[key])

            elif isinstance(original[key], list) and isinstance(incoming[key], list):
                deep_merge_lists(original[key], incoming[key])

            else:
                original[key] = incoming[key]
        else:
            original[key] = incoming[key]

Question 50

If you have an unknown level of dictionaries, then I would suggest a recursive function:

def combineDicts(dictionary1, dictionary2):
    output = {}
    for item, value in dictionary1.iteritems():
        if dictionary2.has_key(item):
            if isinstance(dictionary2[item], dict):
                output[item] = combineDicts(value, dictionary2.pop(item))
        else:
            output[item] = value
    for item, value in dictionary2.iteritems():
         output[item] = value
    return output

Question 51

Overview

The following approach subdivides the problem of a deep merge of dicts into:

A parameterized shallow merge function merge(f)(a,b) that uses a function f to merge two dicts a and b
A recursive merger function f to be used together with merge

Implementation

A function for merging two (non nested) dicts can be written in a lot of ways. I personally like

def merge(f):
    def merge(a,b): 
        keys = a.keys() | b.keys()
        return {key:f(a.get(key), b.get(key)) for key in keys}
    return merge

A nice way of defining an appropriate recursive merger function f is using multipledispatch which allows to define functions that evaluate along different paths depending on the type of their arguments.

from multipledispatch import dispatch

#for anything that is not a dict return
@dispatch(object, object)
def f(a, b):
    return b if b is not None else a

#for dicts recurse 
@dispatch(dict, dict)
def f(a,b):
    return merge(f)(a,b)

Example

To merge two nested dicts simply use merge(f) e.g.:

dict1 = {1:{"a":"A"},2:{"b":"B"}}
dict2 = {2:{"c":"C"},3:{"d":"D"}}
merge(f)(dict1, dict2)
#returns {1: {'a': 'A'}, 2: {'b': 'B', 'c': 'C'}, 3: {'d': 'D'}}

Notes:

The advantages of this approach are:

The function is build from smaller functions that each do a single thing which makes the code simpler to reason about and test
The behaviour is not hard-coded but can be changed and extended as needed which improves code reuse (see example below).

Customization

Some answers also considered dicts that contain lists e.g. of other (potentially nested) dicts. In this case one might want map over the lists and merge them based on position. This can be done by adding another definition to the merger function f:

import itertools
@dispatch(list, list)
def f(a,b):
    return [merge(f)(*arg) for arg in itertools.zip_longest(a, b)]

Question 52

In case someone wants yet another approach to this problem, here’s my solution.

Virtues: short, declarative, and functional in style (recursive, does no mutation).

Potential Drawback: This might not be the merge you’re looking for. Consult the docstring for semantics.

def deep_merge(a, b):
    """
    Merge two values, with `b` taking precedence over `a`.

    Semantics:
    - If either `a` or `b` is not a dictionary, `a` will be returned only if
      `b` is `None`. Otherwise `b` will be returned.
    - If both values are dictionaries, they are merged as follows:
        * Each key that is found only in `a` or only in `b` will be included in
          the output collection with its value intact.
        * For any key in common between `a` and `b`, the corresponding values
          will be merged with the same semantics.
    """
    if not isinstance(a, dict) or not isinstance(b, dict):
        return a if b is None else b
    else:
        # If we're here, both a and b must be dictionaries or subtypes thereof.

        # Compute set of all keys in both dictionaries.
        keys = set(a.keys()) | set(b.keys())

        # Build output dictionary, merging recursively values with common keys,
        # where `None` is used to mean the absence of a value.
        return {
            key: deep_merge(a.get(key), b.get(key))
            for key in keys
        }

Question 53

You could try mergedeep.

Installation

$ pip3 install mergedeep

Usage

from mergedeep import merge

a = {"keyA": 1}
b = {"keyB": {"sub1": 10}}
c = {"keyB": {"sub2": 20}}

merge(a, b, c) 

print(a)
# {"keyA": 1, "keyB": {"sub1": 10, "sub2": 20}}

For a full list of options, check out the docs!

Question 54

There’s a slight problem with andrew cookes answer: In some cases it modifies the second argument b when you modify the returned dict. Specifically it’s because of this line:

if key in a:
    ...
else:
    a[key] = b[key]

If b[key] is a dict, it will simply be assigned to a, meaning any subsequent modifications to that dict will affect both a and b.

a={}
b={'1':{'2':'b'}}
c={'1':{'3':'c'}}
merge(merge(a,b), c) # {'1': {'3': 'c', '2': 'b'}}
a # {'1': {'3': 'c', '2': 'b'}} (as expected)
b # {'1': {'3': 'c', '2': 'b'}} <----
c # {'1': {'3': 'c'}} (unmodified)

To fix this, the line would have to be substituted with this:

if isinstance(b[key], dict):
    a[key] = clone_dict(b[key])
else:
    a[key] = b[key]

Where clone_dict is:

def clone_dict(obj):
    clone = {}
    for key, value in obj.iteritems():
        if isinstance(value, dict):
            clone[key] = clone_dict(value)
        else:
            clone[key] = value
    return

Still. This obviously doesn’t account for list, set and other stuff, but I hope it illustrates the pitfalls when trying to merge dicts.

And for completeness sake, here is my version, where you can pass it multiple dicts:

def merge_dicts(*args):
    def clone_dict(obj):
        clone = {}
        for key, value in obj.iteritems():
            if isinstance(value, dict):
                clone[key] = clone_dict(value)
            else:
                clone[key] = value
        return

    def merge(a, b, path=[]):
        for key in b:
            if key in a:
                if isinstance(a[key], dict) and isinstance(b[key], dict):
                    merge(a[key], b[key], path + [str(key)])
                elif a[key] == b[key]:
                    pass
                else:
                    raise Exception('Conflict at `{path}\''.format(path='.'.join(path + [str(key)])))
            else:
                if isinstance(b[key], dict):
                    a[key] = clone_dict(b[key])
                else:
                    a[key] = b[key]
        return a
    return reduce(merge, args, {})

Question 55

This version of the function will account for N number of dictionaries, and only dictionaries — no improper parameters can be passed, or it will raise a TypeError. The merge itself accounts for key conflicts, and instead of overwriting data from a dictionary further down the merge chain, it creates a set of values and appends to that; no data is lost.

It might not be the most effecient on the page, but it’s the most thorough and you’re not going to lose any information when you merge your 2 to N dicts.

def merge_dicts(*dicts):
    if not reduce(lambda x, y: isinstance(y, dict) and x, dicts, True):
        raise TypeError, "Object in *dicts not of type dict"
    if len(dicts) < 2:
        raise ValueError, "Requires 2 or more dict objects"


    def merge(a, b):
        for d in set(a.keys()).union(b.keys()):
            if d in a and d in b:
                if type(a[d]) == type(b[d]):
                    if not isinstance(a[d], dict):
                        ret = list({a[d], b[d]})
                        if len(ret) == 1: ret = ret[0]
                        yield (d, sorted(ret))
                    else:
                        yield (d, dict(merge(a[d], b[d])))
                else:
                    raise TypeError, "Conflicting key:value type assignment"
            elif d in a:
                yield (d, a[d])
            elif d in b:
                yield (d, b[d])
            else:
                raise KeyError

    return reduce(lambda x, y: dict(merge(x, y)), dicts[1:], dicts[0])

print merge_dicts({1:1,2:{1:2}},{1:2,2:{3:1}},{4:4})

output: {1: [1, 2], 2: {1: 2, 3: 1}, 4: 4}

Question 56

Since dictviews support set operations, I was able to greatly simplify jterrace’s answer.

def merge(dict1, dict2):
    for k in dict1.keys() - dict2.keys():
        yield (k, dict1[k])

    for k in dict2.keys() - dict1.keys():
        yield (k, dict2[k])

    for k in dict1.keys() & dict2.keys():
        yield (k, dict(merge(dict1[k], dict2[k])))

Any attempt to combine a dict with a non dict (technically, an object with a ‘keys’ method and an object without a ‘keys’ method) will raise an AttributeError. This includes both the initial call to the function and recursive calls. This is exactly what I wanted so I left it. You could easily catch an AttributeErrors thrown by the recursive call and then yield any value you please.

Question 57

Short-n-sweet:

from collections.abc import MutableMapping as Map

def nested_update(d, v):
"""
Nested update of dict-like 'd' with dict-like 'v'.
"""

for key in v:
    if key in d and isinstance(d[key], Map) and isinstance(v[key], Map):
        nested_update(d[key], v[key])
    else:
        d[key] = v[key]

This works like (and is build on) Python’s dict.update method. It returns None (you can always add return d if you prefer) as it updates dict d in-place. Keys in v will overwrite any existing keys in d (it does not try to interpret the dict’s contents).

It will also work for other (“dict-like”) mappings.

Question 58

The code will depend on your rules for resolving merge conflicts, of course. Here’s a version which can take an arbitrary number of arguments and merges them recursively to an arbitrary depth, without using any object mutation. It uses the following rules to resolve merge conflicts:

dictionaries take precedence over non-dict values ({"foo": {...}} takes precedence over {"foo": "bar"})
later arguments take precedence over earlier arguments (if you merge {"a": 1}, {"a", 2}, and {"a": 3} in order, the result will be {"a": 3})

try:
    from collections import Mapping
except ImportError:
    Mapping = dict

def merge_dicts(*dicts):                                                            
    """                                                                             
    Return a new dictionary that is the result of merging the arguments together.   
    In case of conflicts, later arguments take precedence over earlier arguments.   
    """                                                                             
    updated = {}                                                                    
    # grab all keys                                                                 
    keys = set()                                                                    
    for d in dicts:                                                                 
        keys = keys.union(set(d))                                                   

    for key in keys:                                                                
        values = [d[key] for d in dicts if key in d]                                
        # which ones are mapping types? (aka dict)                                  
        maps = [value for value in values if isinstance(value, Mapping)]            
        if maps:                                                                    
            # if we have any mapping types, call recursively to merge them          
            updated[key] = merge_dicts(*maps)                                       
        else:                                                                       
            # otherwise, just grab the last value we have, since later arguments    
            # take precedence over earlier arguments                                
            updated[key] = values[-1]                                               
    return updated

Question 59

I had two dictionaries (a and b) which could each contain any number of nested dictionaries. I wanted to recursively merge them, with b taking precedence over a.

Considering the nested dictionaries as trees, what I wanted was:

To update a so that every path to every leaf in b would be represented in a
To overwrite subtrees of a if a leaf is found in the corresponding path in b
- Maintain the invariant that all b leaf nodes remain leafs.

The existing answers were a little complicated for my taste and left some details on the shelf. I hacked together the following, which passes unit tests for my data set.

  def merge_map(a, b):
    if not isinstance(a, dict) or not isinstance(b, dict):
      return b

    for key in b.keys():
      a[key] = merge_map(a[key], b[key]) if key in a else b[key]
    return a

Example (formatted for clarity):

 a = {
    1 : {'a': 'red', 
         'b': {'blue': 'fish', 'yellow': 'bear' },
         'c': { 'orange': 'dog'},
    },
    2 : {'d': 'green'},
    3: 'e'
  }

  b = {
    1 : {'b': 'white'},
    2 : {'d': 'black'},
    3: 'e'
  }


  >>> merge_map(a, b)
  {1: {'a': 'red', 
       'b': 'white',
       'c': {'orange': 'dog'},},
   2: {'d': 'black'},
   3: 'e'}

The paths in b that needed to be maintained were:

1 -> 'b' -> 'white'
2 -> 'd' -> 'black'
3 -> 'e'.

a had the unique and non-conflicting paths of:

1 -> 'a' -> 'red'
1 -> 'c' -> 'orange' -> 'dog'

so they are still represented in the merged map.

Question 60

I have an iterative solution – works much much better with big dicts & a lot of them (for example jsons etc):

import collections


def merge_dict_with_subdicts(dict1: dict, dict2: dict) -> dict:
    """
    similar behaviour to builtin dict.update - but knows how to handle nested dicts
    """
    q = collections.deque([(dict1, dict2)])
    while len(q) > 0:
        d1, d2 = q.pop()
        for k, v in d2.items():
            if k in d1 and isinstance(d1[k], dict) and isinstance(v, dict):
                q.append((d1[k], v))
            else:
                d1[k] = v

    return dict1

note that this will use the value in d2 to override d1, in case they are not both dicts. (same as python’s dict.update())

some tests:

def test_deep_update():
    d = dict()
    merge_dict_with_subdicts(d, {"a": 4})
    assert d == {"a": 4}

    new_dict = {
        "b": {
            "c": {
                "d": 6
            }
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d == {
        "a": 4,
        "b": {
            "c": {
                "d": 6
            }
        }
    }

    new_dict = {
        "a": 3,
        "b": {
            "f": 7
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d == {
        "a": 3,
        "b": {
            "c": {
                "d": 6
            },
            "f": 7
        }
    }

    # test a case where one of the dicts has dict as value and the other has something else
    new_dict = {
        'a': {
            'b': 4
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d['a']['b'] == 4

I’ve tested with around ~1200 dicts – this method took 0.4 seconds, while the recursive solution took ~2.5 seconds.

Question 61

This should help in merging all items from dict2 into dict1:

for item in dict2:
    if item in dict1:
        for leaf in dict2[item]:
            dict1[item][leaf] = dict2[item][leaf]
    else:
        dict1[item] = dict2[item]

Please test it and tell us whether this is what you wanted.

EDIT:

The above mentioned solution merges only one level, but correctly solves the example given by OP. To merge multiple levels, the recursion should be used.

Question 62

I’ve been testing your solutions and decided to use this one in my project:

def mergedicts(dict1, dict2, conflict, no_conflict):
    for k in set(dict1.keys()).union(dict2.keys()):
        if k in dict1 and k in dict2:
            yield (k, conflict(dict1[k], dict2[k]))
        elif k in dict1:
            yield (k, no_conflict(dict1[k]))
        else:
            yield (k, no_conflict(dict2[k]))

dict1 = {1:{"a":"A"}, 2:{"b":"B"}}
dict2 = {2:{"c":"C"}, 3:{"d":"D"}}

#this helper function allows for recursion and the use of reduce
def f2(x, y):
    return dict(mergedicts(x, y, f2, lambda x: x))

print dict(mergedicts(dict1, dict2, f2, lambda x: x))
print dict(reduce(f2, [dict1, dict2]))

Passing functions as parameteres is key to extend jterrace solution to behave as all the other recursive solutions.

Question 63

Easiest way i can think of is :

#!/usr/bin/python

from copy import deepcopy
def dict_merge(a, b):
    if not isinstance(b, dict):
        return b
    result = deepcopy(a)
    for k, v in b.iteritems():
        if k in result and isinstance(result[k], dict):
                result[k] = dict_merge(result[k], v)
        else:
            result[k] = deepcopy(v)
    return result

a = {1:{"a":'A'}, 2:{"b":'B'}}
b = {2:{"c":'C'}, 3:{"d":'D'}}

print dict_merge(a,b)

Output:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}

Question 64

I have another slightly different solution here:

def deepMerge(d1, d2, inconflict = lambda v1,v2 : v2) :
''' merge d2 into d1. using inconflict function to resolve the leaf conflicts '''
    for k in d2:
        if k in d1 : 
            if isinstance(d1[k], dict) and isinstance(d2[k], dict) :
                deepMerge(d1[k], d2[k], inconflict)
            elif d1[k] != d2[k] :
                d1[k] = inconflict(d1[k], d2[k])
        else :
            d1[k] = d2[k]
    return d1

By default it resolves conflicts in favor of values from the second dict, but you can easily override this, with some witchery you may be able to even throw exceptions out of it. :).

问题：如何使用python识别文件是普通文件还是目录

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问题：类方法生成“ TypeError：…为关键字参数获得了多个值……”

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解

附加信息

Solution

additional information

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问题：使用argparse获取选定的子命令

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问题：在python中继承和覆盖__init__

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问题：您如何在python中检查字符串是否仅包含数字？

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问题：如何合并字典的字典？

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最简单的情况：“叶是嵌套的字典，以空字典结尾”：

复杂的情况：“叶子是任何其他类型：”

适用于原始问题：

Simplest Case: “leaves are nested dicts that end in empty dicts”:

Complex Case: “leaves are of any other type:”

Application to the original question:

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问题：在python中继承和覆盖init